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# Distribution of 0 Digits in Factorial Series

Distribution of 0 Digits in Factorial Series is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Large Factorials and the Distribution of '0' in base 10 digits.

About the task

We can see that some features of factorial numbers (the series of numbers 1!, 2!, 3!, ...) come about because such numbers are the product of a series of counting numbers, and so those products have predictable factors. For example, all factorials above 1! are even numbers, since they have 2 as a factor. Similarly, all factorials from 5! up end in a 0, because they have 5 and 2 as factors, and thus have 10 as a factor. In fact, the factorial integers add another 0 at the end of the factorial for every step of 5 upward: 5! = 120, 10! = 3628800, 15! = 1307674368000, 16! = 20922789888000 and so on.

Because factorial numbers, which quickly become quite large, continue to have another terminal 0 on the right hand side of the number for every factor of 5 added to the factorial product, one might think that the proportion of zeros in a base 10 factorial number might be close to 1/5. However, though the factorial products add another terminating 0 every factor of 5 multiplied into the product, as the numbers become quite large, the number of digits in the factorial product expands exponentially, and so the number above the terminating zeros tends toward 10% of each digit from 0 to 1 as the factorial becomes larger. Thus, as the factorials become larger, the proportion of 0 digits in the factorial products shifts slowly from around 1/5 toward 1/10, since the number of terminating zeros in n! increases only in proportion to n, whereas the number of digits of n! in base 10 increases exponentially.

The task

Create a function to calculate the mean of the proportions of 0 digits out of the total digits found in each factorial product from 1! to N!. This proportion of 0 digits in base 10 should be calculated using the number as printed as a base 10 integer.

Example: for 1 to 6 we have 1!, 2!, 3!, 4!, 5!, 6!, or (1, 2, 6, 24, 120, 720), so we need the mean of (0/1, 0/1, 0/1, 0/2, 1/3, 1/3) = (2/3) (totals of each proportion) / 6 (= N), or 0.1111111...

Example: for 1 to 25 the mean of the proportions of 0 digits in the factorial products series of N! with N from 1 to 25 is 0.26787.

Do this task for 1 to N where N is in (100, 1000, and 10000), so, compute the mean of the proportion of 0 digits for each product in the series of each of the factorials from 1 to 100, 1 to 1000, and 1 to 10000.

Stretch task

Find the N in 10000 < N < 50000 where the mean of the proportions of 0 digits in the factorial products from 1 to N permanently falls below 0.16. This task took many hours in the Python example, though I wonder if there is a faster algorithm out there.

## Go

Library: Go-rcu

Brute force as I'll be surprised if there is a faster 'exact' algorithm for this task.

However, the combination of a native code compiler and GMP really cuts down the times (2.8 seconds for the basic task and 182.5 seconds for the stretch goal). Expect these times to be reduced further by the fastest languages.

`package main import (    "fmt"    big "github.com/ncw/gmp"    "rcu") func main() {    fact  := big.NewInt(1)    sum   := 0.0    first := int64(0)    firstRatio := 0.0        fmt.Println("The mean proportion of zero digits in factorials up to the following are:")    for n := int64(1); n <= 50000; n++  {        fact.Mul(fact, big.NewInt(n))        bytes  := []byte(fact.String())        digits := len(bytes)        zeros  := 0        for _, b := range bytes {            if b == '0' {                zeros++            }        }        sum += float64(zeros)/float64(digits)        ratio := sum / float64(n)        if n == 100 || n == 1000 || n == 10000 {            fmt.Printf("%6s = %12.10f\n", rcu.Commatize(int(n)), ratio)        }         if first > 0 && ratio >= 0.16 {            first = 0            firstRatio = 0.0        } else if first == 0 && ratio < 0.16 {            first = n            firstRatio = ratio                   }    }    fmt.Printf("%6s = %12.10f", rcu.Commatize(int(first)), firstRatio)    fmt.Println(" (stays below 0.16 after this)")    fmt.Println(sum/ 50000)}`
Output:
```The mean proportion of zero digits in factorials up to the following are:
100 = 0.2467531862
1,000 = 0.2035445511
10,000 = 0.1730038482
47,332 = 0.1599999958 (stays below 0.16 after this)
```

## Julia

`function meanfactorialdigits(N, goal = 0.0)    factoril, proportionsum = big"1", 0.0    for i in 1:N        factoril *= i        d = digits(factoril)        zero_proportion_in_fac = count(x -> x == 0, d) / length(d)        proportionsum += zero_proportion_in_fac        propmean = proportionsum / i        if i > 15 && propmean <= goal            println("The mean proportion dips permanently below \$goal at \$i.")            break        end        if i == N            println("Mean proportion of zero digits in factorials to \$N is ", propmean)        end    endend @time foreach(meanfactorialdigits, [100, 1000, 10000]) @time meanfactorialdigits(50000, 0.16) `
Output:
```Mean proportion of zero digits in factorials to 100 is 0.24675318616743216
Mean proportion of zero digits in factorials to 1000 is 0.20354455110316458
Mean proportion of zero digits in factorials to 10000 is 0.17300384824186707
3.030182 seconds (297.84 k allocations: 1.669 GiB, 0.83% gc time, 0.28% compilation time)
The mean proportion dips permanently below 0.16 at 47332.
179.157788 seconds (3.65 M allocations: 59.696 GiB, 1.11% gc time)
```

## Nim

### Task

Library: bignum
`import strutils, std/monotimesimport bignum let t0 = getMonoTime()var sum = 0.0var f = newInt(1)var lim = 100for n in 1..10_000:  f *= n  let str = \$f  sum += str.count('0') / str.len  if n == lim:    echo n, ":\t", sum / float(n)    lim *= 10echo()echo getMonoTime() - t0`
Output:
```100:    0.2467531861674322
1000:   0.2035445511031646
10000:  0.1730038482418671

(seconds: 2, nanosecond: 857794404)```

### Stretch task

Library: bignum

At each step, we eliminate the trailing zeroes to reduce the length of the number and save some time. But this is not much, about 8%.

`import strutils, std/monotimesimport bignum let t0 = getMonoTime()var sum = 0.0var first = 0var f = newInt(1)var count0 = 0for n in 1..<50_000:  f *= n  while f mod 10 == 0:    # Reduce the length of "f".    f = f div 10    inc count0  let str = \$f  sum += (str.count('0') + count0) / (str.len + count0)  if sum / float(n) < 0.16:    if first == 0: first = n  else:    first = 0 echo "Permanently below 0.16 at n = ", firstecho "Execution time: ", getMonoTime() - t0`
Output:
```Permanently below 0.16 at n = 47332
Execution time: (seconds: 190, nanosecond: 215845101)```

## Pascal

Doing the calculation in Base 1,000,000,000 like in Primorial_numbers#alternative.
The most time consuming is converting to string and search for zeros.
Therefor I do not convert to string.I divide the base in sections of 3 digits with counting zeros in a lookup table.

`program Factorial;{\$IFDEF FPC} {\$MODE DELPHI} {\$Optimization ON,ALL} {\$ENDIF}uses  sysutils;type  tMul = array of LongWord;  tpMul = pLongWord;const  LongWordDec = 1000*1000*1000;  LIMIT = 50000;var  CountOfZero : array[0..999] of byte;  SumOfRatio :array[0..LIMIT] of extended;  procedure OutMul(pMul:tpMul;Lmt :NativeInt);// for testingBegin  write(pMul[lmt]);  For lmt := lmt-1  downto 0 do    write(Format('%.9d',[pMul[lmt]]));  writeln;end; procedure InitCoZ;//Init Lookup table for 3 digitsvar  x,y : integer;begin  fillchar(CountOfZero,SizeOf(CountOfZero),#0);  CountOfZero := 3; //000  For x := 1 to 9 do  Begin    CountOfZero[x] := 2;     //00x    CountOfZero[10*x] := 2;  //0x0    CountOfZero[100*x] := 2; //x00    y := 10;    repeat      CountOfZero[y+x] := 1;      //0yx      CountOfZero[10*y+x] := 1;   //y0x      CountOfZero[10*(y+x)] := 1; //yx0      inc(y,10)    until y > 100;  end;end; function getFactorialDecDigits(n:NativeInt):NativeInt;var  res: extended;Begin  result := -1;  IF (n > 0) AND (n <= 1000*1000) then  Begin    res := 0;    repeat res := res+ln(n); dec(n); until n < 2;    result := trunc(res/ln(10))+1;  end;end; function CntZero(pMul:tpMul;Lmt :NativeInt):NativeUint;//count zeros in Base 1,000,000,000 numbervar  q,r : LongWord;  i : NativeInt;begin  result := 0;  For i := Lmt-1 downto 0 do  Begin    q := pMul[i];    r := q DIV 1000;    result +=CountOfZero[q-1000*r];//q-1000*r == q mod 1000    q := r;    r := q DIV 1000;    result +=CountOfZero[q-1000*r];    q := r;    r := q DIV 1000;    result +=CountOfZero[q-1000*r];  end;//special case first digits no leading '0'  q := pMul[lmt];  while q >= 1000 do  begin    r := q DIV 1000;    result +=CountOfZero[q-1000*r];    q := r;  end;  while q > 0 do  begin    r := q DIV 10;    result += Ord( q-10*r= 0);    q := r;  end;end; function GetCoD(pMul:tpMul;Lmt :NativeInt):NativeUint;//count of decimal digitsvar  i : longWord;begin  result := 9*Lmt;  i := pMul[Lmt];  while i > 1000 do  begin    i := i DIV 1000;    inc(result,3);  end;  while i > 0 do  begin    i := i DIV 10;    inc(result);  end;end; procedure DoChecks(pMul:tpMul;Lmt,i :NativeInt);//(extended(1.0)* makes TIO.RUN faster // only using FPU?Begin  SumOfRatio[i] := SumOfRatio[i-1] + (extended(1.0)*CntZero(pMul,Lmt))/GetCoD(pMul,Lmt);end; function MulByI(pMul:tpMul;UL,i :NativeInt):NativeInt;var  prod  : Uint64;  j     : nativeInt;  carry : LongWord;begin  result := UL;  carry := 0;  For j := 0 to result do  Begin    prod  := i*pMul+Carry;    Carry := prod Div LongWordDec;    pMul := Prod - LongWordDec*Carry;    inc(pMul);  end;   IF Carry <> 0 then  Begin    inc(result);    pMul:= Carry;  End;end; procedure getFactorialExact(n:NativeInt);var  MulArr : tMul;  pMul : tpMul;  i,ul : NativeInt;begin  i := getFactorialDecDigits(n) DIV 9 +10;  Setlength(MulArr,i);  pMul := @MulArr;  Ul := 0;  pMul[Ul]:= 1;  i := 1;  repeat    UL := MulByI(pMul,UL,i);    //Now do what you like to do with i!    DoChecks(pMul,UL,i);    inc(i);  until i> n;end; procedure Out_(i: integer);begin  if i > LIMIT then    EXIT;  writeln(i:8,SumOfRatio[i]/i:18:15);end; var  i : integer;Begin  InitCoZ;  SumOfRatio:= 0;  getFactorialExact(LIMIT);  Out_(100);  Out_(1000);  Out_(10000);  Out_(50000);  i := limit;  while i >0 do  Begin    if SumOfRatio[i]/i >0.16 then      break;    dec(i);  end;  inc(i);  writeln('First ratio < 0.16 ', i:8,SumOfRatio[i]/i:20:17);end.`
Output:
```     100 0.246753186167432
1000 0.203544551103165
10000 0.173003848241866
50000 0.159620054602269
First ratio < 0.16    47332 0.15999999579985665
Real time: 4.898 s  CPU share: 99.55 % // 2.67s on 2200G freepascal 3.2.2```

## Python

`def facpropzeros(N, verbose = True):    proportions = [0.0] * N    fac, psum = 1, 0.0    for i in range(N):        fac *= i + 1        d = list(str(fac))        psum += sum(map(lambda x: x == '0', d)) / len(d)        proportions[i] = psum / (i + 1)     if verbose:        print("The mean proportion of 0 in factorials from 1 to {} is {}.".format(N, psum / N))     return proportions  for n in [100, 1000, 10000]:    facpropzeros(n) props = facpropzeros(47500, False)n = (next(i for i in reversed(range(len(props))) if props[i] > 0.16)) print("The mean proportion dips permanently below 0.16 at {}.".format(n + 2)) `
Output:
```The mean proportion of 0 in factorials from 1 to 100 is 0.24675318616743216.
The mean proportion of 0 in factorials from 1 to 1000 is 0.20354455110316458.
The mean proportion of 0 in factorials from 1 to 10000 is 0.17300384824186707.
The mean proportion dips permanently below 0.16 at 47332.
```

The means can be plotted, showing a jump from 0 to over 0.25, followed by a slowly dropping curve:

`import matplotlib.pyplot as pltplt.plot([i+1 for i in range(len(props))], props) `

## Raku

Works, but depressingly slow for 10000.

`sub postfix:<!> (Int \$n) { ( constant factorial = 1, 1, |[\*] 2..* )[\$n] }sink 10000!; # prime the iterator to allow multithreading sub zs (\$n) { ( constant zero-share = (^Inf).race(:32batch).map: { (.!.comb.Bag){'0'} / .!.chars } )[\$n] } .say for (     100    ,1000    ,10000).map:  -> \n { "{n}: {([+] (^n).map: *.&zs) / n}" }`
Output:
```100: 0.24485445199021696
1000: 0.20336075048011162
10000: 0.17298757510670162
```

## REXX

`/*REXX program computes the mean of the proportion of "0" digits a series of factorials.*/parse arg \$                                      /*obtain optional arguments from the CL*/if \$='' | \$=","  then \$= 100 1000 10000          /*not specified?  Then use the default.*/#= words(\$)                                      /*the number of ranges to be used here.*/numeric digits 100                               /*increase dec. digs, but only to 100. */big= word(\$, #);  != 1                           /*obtain the largest number in ranges. */                                do i=1  for big  /*calculate biggest  !  using 100 digs.*/                                != ! * i         /*calculate the factorial of  BIG.     */                                end   /*i*/if pos('E', !)>0  then do                        /*In exponential format?  Then get EXP.*/                       parse var !  'E'  x       /*parse the exponent from the number.  */                       numeric digits    x+1     /*set the decimal digits to  X  plus 1.*/                       end                       /* [↑]  the  +1  is for the dec. point.*/ title= ' mean proportion of zeros in the (decimal) factorial products for  N'say '     N     │'center(title, 80)              /*display the title for the output.    */say '───────────┼'center(""   , 80, '─')         /*   "     a   sep   "   "     "       */   do j=1  for #;  n= word(\$, j)                  /*calculate some factorial ranges.     */  say center( commas(n), 11)'│' left(0dist(n), 75)...    /*show results for above range.*/  end   /*j*/ say '───────────┴'center(""   , 80, '─')         /*display a foot sep for the output.   */exit 0                                           /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/commas: parse arg ?;  do jc=length(?)-3  to 1  by -3; ?=insert(',', ?, jc); end;  return ?/*──────────────────────────────────────────────────────────────────────────────────────*/0dist:  procedure; parse arg z;        != 1;         y= 0                     do k=1  for z;    != ! * k;     y= y   +   countstr(0, !) / length(!)                     end   /*k*/        return y/z`
output   when using the default inputs:
```     N     │       mean proportion of zeros in the (decimal) factorial products for  N
───────────┼────────────────────────────────────────────────────────────────────────────────
100    │ 0.2467531861674322177784158871973526991129407033266153063813195937196095976...
1,000   │ 0.2035445511031646356400438031711455302985741167890402203486699704599684047...
10,000   │ 0.1730038482418660531800366428930706156810278809057883361518852958446868172...
───────────┴────────────────────────────────────────────────────────────────────────────────
```

## Wren

Library: Wren-big
Library: Wren-fmt

Very slow indeed, 10.75 minutes to reach N = 10,000.

`import "/big" for BigIntimport "/fmt" for Fmt var fact = BigInt.onevar sum = 0System.print("The mean proportion of zero digits in factorials up to the following are:")for (n in 1..10000) {    fact = fact * n    var bytes = fact.toString.bytes    var digits = bytes.count    var zeros  = bytes.count { |b| b == 48 }    sum = sum + zeros / digits    if (n == 100 || n == 1000 || n == 10000) {        Fmt.print("\$,6d = \$12.10f", n, sum / n)    }}`
Output:
```The mean proportion of zero digits in factorials up to the following are:
100 = 0.2467531862
1,000 = 0.2035445511
10,000 = 0.1730038482
```